Inductance Circuit, Electrostatics, Physics Assignment Help

Let a source of alternating be connected to a circuit containing a pure inductance L only suppose the alternating supplied is represented by

E = E0 sin wt

If si/st is the rebate of change of current through L at any instant then induced in the inductor at the same instant is - L si/st. the negative sign indicates that induced opposes the change of current.

Usually inductors have some resistance in their windings. By we shall assume that resistance of this inductor is negligible. The circuit is therefore a purely inductive circuit. To maintain the flow of current in this circuit applied voltage the flow of current in this circuit applied voltage must be equal and opposite to the induced voltage.E = - (- L dI/dt) = E0 sin wt

Or dI = E0 / L sin wt dt

Integrating both sides we get

I = E0 / L ∫sin w t dt

I = E0/L (- cos wt/w) + constant

The integration constant has the dimensions of current, and is time-independent. As of the source oscillates symmetrically about zero, the current it sustains must also oscillate symmetrically about zero. Therefore no component of current which is time indecent exists. Therefore integration constant is zero.

∴ form I = (- E0/wL) cos wt

I = (- E0/wL) [sin (π/2)] – wt

I= E0 / wL sin [wt – (π/2)

The current will be maximum I = Iw. When sin (wt = π/2) = maximum = 1.

From above eqn, I0 = (E0 /wL) × 1

Putting in (eq.), we getI = I0 sin [wt – (π/2)]

This is the form of alternating current developed.

Comparing with we find that in an circuit containing L only alternating current I lags behind the alternating voltage E by a phase Only of 90° by one fourth of a period.

Conversely voltage across L leads the current by a (c) phase angle of 90° this shown in

Figure (b) represents the vector diagram or the phasor diagram of circuit containing L only. The vector representing E0 makes an angle (wt) with OX, As current lags behind the by 90° therefore, phasor representing Io is turned clock wise through 90° from the direction of E0 the projections of these phasor on YOY give instantaneous values E and I as shown then E0 and Io rotate with frequency w curves in (c) are generated.

Comparing eqn. with Ohm’s law equation viz. current = voltage / resistance we find that (wL) represents the effective resistance offered by inductance L. this is called inductive reactance and is denoted by X,.

Thus , XL = wL = 2π vL, where v is the frequency of supply.

The inductive reactance limits the current in a purely inductive circuit in the same way as resistance limits the current in a purely resistance circuit. Clearly the inductive reactance (XL) is directly proportional to the inductance (L) and to the frequency (v) of the current.

In dc circuit v = 0 ∴ XL = 0

XL = w L » 1 / sec (Henry) = 1 / sec volt amp/ sec = ohm.

Hence XL is measured in ohm just as resistance is measured in ohm. The dimensions of inductive reactance are the same as those of resistance.

It showily be clearly understood, that in a circuit containing, R opposition arises on account of obstruction to the passage of electrons through are resistor. In an inductive circuit it is the self induced that opposes the growth of current.

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